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Submitted by Hemicarka5 on Mon, 12/01/2014 - 23:16
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Koliko cm3 rastvora broma u ugljentetrahloridu (160g brom/1 dm3) moze obezbijediti 0,84g heksena?

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milicaradenkovic - Tue, 12/02/2014 - 07:29

C6H12+Br2/CCl4------------------>C6H12Br2

nC6H12)=m/Mr=0.84/84=0.01 mol

n(Br2)=n(C6H12)=0.01 mol

c(Br2)=y/Mr=160/160=1 mol/l

V(Br2)=n(Br2)/c(Br2)=0.01/1=0.01 l=10 ml